Congruences for the class numbers of real cyclic sextic number fields

V O ~ . 42 NO. 10 SCIENCE IN CHINA (Series A) October 1999 Congruences for the class numbers of real cyclic sextic number fields * LIU Tong ($1 a ) ** ( Department of Applied Mathematics, Tsinghua University, Beijing 100084, China) Received September 21, 1998 Abstract Ifit K, be a real cyclic sextic number field, and K , , K3 its quadratic and cubic subfield. Let h( L ) de- note the ideal class number of field L . Seven congruences for h - = h ( K6)/( h( K2) h( K3) ) are obtained. In particu- lar, when the conductor f6 of K6 is a prime p , Ch - = B e B s w (mod p ) , where C is an explicitly given constant, and 8, is the Bernoulli number. These results on real cyclic sextic fields are an extension of the results on quadratic and cyclic quartic fields. Keywords: real cyclic sextic number field, class number, p-adic L-betion. Ankeny et al. obtained several congruences for the class numbers h of real quadratic fields k [ l 3 , some of which were also obtained by Kiselev. If the discriminant of k is a prime number p E 1 (mod 4) and e0 = ( t + u &)/2 is the fundamental unit of k , then h u / t B d ( m o d p ) , 2 (0.1) where B, is the nth Bernoulli number. Further results on more general fields k were obtained later by Carlitz, Slavutskii , Lang and Schertz , and by LU['] . In 1982 and 1987, , zhangE4] obtained analogue of (0.1 ) for the cyclic cubic fields respectively. Furthermore, in ref. [5] , Zhang consid- ered similar problem for cyclic quartic fields, and obtained ten congruences for the class numbers. In this paper we obtain seven congruences for the class numbers of the real cyclic sextic fields. Now, let K6 be a real cyclic number field of degree six over the rationals 0 and f6 its conductor. Let K2, K3 be quadratic and cubic subfield of K6 and f 2 , f3 their conductors respectively, and f * = ( fi , f3) the greatest common divisor for f 2 , f3. Denote the Galois group of K6 over by Gal( K6/Q) = ( a ) , and the character group of K6 by ( X ) . Let U6, U3, U2 denote unit groups of K6 , Kg, K2 respectively; let p , & be fundamental units of K3,K2respectively; that is U g = ( - l , p , p " ) , U2= ( - 1 , ~ ) . Let UR = 1 % E U6 I Ng/2 (~ ) = f 1 , N 6 / 3 ( ~ ) = f 11 be the relative unit group of Kg, where N6/2, N6/3 are norm maps from K6 to K2, K3 respectively. From ref. [6] we know that there exists a relative unit E E UR such that UR = ( - 1) x ( E ) x ( E " ) . + Project supported by the National Natural Science Foundation of China (Grant No. 19771052) . ** Current address : Department of Applied Mathematics, Ocean University of Qingdao, Qingdao 266003, China. I010 SCIENCE IN CHINA (Series A ) Vol. 42 Then E is said to be the fundamental relative unit. h l l m a 1. Let u " = U2 U3 U , . Then U * is finite index subgroup of U6 . Let Q = ( U6 : U * ) . Then Q = 1 2 / Q , Q 2 , where Q , = ( U 2 : N 6 , 2 ( U 6 > ) ; Q 2 = ( ~ 3 : ( - 1 ) ~ 6 , 3 ( ~ 6 ) ) ; and Q I = 1 , 4 ; Q 2 = 1 , 3 ; Q = 1 ,3 ,4 ,12 . Lemma 2. We can write the fundamental unit E as - - - 0 2 - E = x + yr + yr + J ( x + yr + y r ) 2 - 1 = x + yr + yr + ,-(XI + y'r + ~ ' r ) , f wherex,x l € ,, y , y ' E t 3 ( F 3 ) , arep-adical ly integralforp#2,3, r = X 2 ( a ) { ; is a(& No. 10 CONGRUENCES FOR CLASS NUMBERS 101 1 and the choice of symbols + is according to N6/2( E ) = + 1. 0 2 Corollary. ZfX,=x+-x '= * l ( m o d p ) , t h e n x ~ * l ( m o d p ) , p l x ' , and .f * f 2 Theorem 5 . Suppose that p 1 - p # 3 ,2 . Then we have f " ' h - ( ~ , ) 1 8 ( x ' - 1 + 2j$f3)/f6 = QBp-l ,x,! Bp-l 2 ,XI,! (mod p 1. 3 In the following, we let R; = - ( ~ O ~ ~ E ~ ' ) ~ , where logp denotes the p-adic logarithm. Q i r , Theorem 6 . Supposing that V f 6 , p 2 5, we have Bp-l B d , 2 ,X,! 2 ' X d (mod p ) . P Remark. It can be expected that congruence in Theorem 1 determines h - ( K6) uniquely, since l imh- ( ~ ~ ) / p ' + € = 0 P- " ( 0 . 2 ) for arbitrary E > 0 in this case ( x ~ here means x / y+O(p+O) ) . In fact, by the Siegel-Brawer theorem we have 1 lim log(hR)/log I D I = 5, I D I - m ( 0 . 3 ) where the limit is taken over the fields K of degree n , D is its discriminant, h its class number, and R its regulator. Thus, for the cyclic sextic fields K6 and its subfields K2, K 3 , we have h ( Kg) R6< 1 1 1 I D61Z", I D 2 1 ? - ' < h ( K 2 ) R 2 a n d I D 3 I T - € < h ( K 3 ) R 3 . Then we have h - ( K 6 ) R - < I D - fore, (0 .2) follows from the fact that R - 3 M"' , where M is positve constant. 1 Structure of cyclic sextic fields We need to describe the cyclic sextic field Kg by its charater group k6 = ( X ) . Since the charac- 2 5 2 3 ter groups of K2, K3 are ( x 3 ) , ( X ) respectively and ~6 = ( X ) is generated by X = X X , let the decomposition of the character X be X = +t60...6L, (1 .1 ) where #i is a pure sextic character, whose conductor is a prime p i , p i = 1 (mod 6) , i = 1, ... , s ; is a cubic character, whose conductor is a prime qj , qj = 1 (mod 3 ) , j = 1, - - - , t ; ek is a quadratic char- acter, whose conductor is a prime rk , rk = 1 (mod 2 ) , k = 1 , ... , I ; go, Bo correspond to characters whose conductors are 3" and 2P , respectively. Here CI = 0 , 1 , 2 , P = 0 , 2 , 3 . If a = 1 , +o is a 1012 SCIENCE IN CHINA (Series A ) Vol. 42 quadratic character, while a = 2 , go is a cubic character. When /3 > 0 , Bo is always a quadratic character. Thus, the conductor f6 of K6 is f6 = 283apl."psq1'. 'qt~l"' ' l . ( I . z ) If P = 1 , then f2 = 2P3rl-. .rlpl.- .p,, f3 = pl..-p,ql-.-q,, f * = (f2, f3) = pl*..p,. If p # l , then f2=29,---rlp1.--ps,f3=3apl---pSq1.--qt, f X = ( f 2 , f 3 ) = ~ 1 . . . ~ s . - p d 5 ( p i - 1) #i is a sextic character and its conductor is p i , so Qi = w 6 or w 6 = w 6 for pi pi Pi If a = 2 , then O 0 ( y ) = (9) ; if a - 3 , then 19 , (~) = third mot of unity. Thus, K6 has 2" - ' ( P # 2) or 2" ' ( P = 2) choices according to the conductor f6 in (1 .2 ) 2 p-adic L-functions and their applications 'L2 We use the notations of ref. [7] and choose an embedding C / Q into C p / Q P such that gi = up6 e=_l (or $ j = u p 3 ) when p = pi(or qj) for i = 1;-.,s(or j = I , . . . , t ) . For any Dirichlet character X and any prime number p , there exists a p-adic L-function, such that - n Lp(l - n , ~ ) = - ( 1 - Xw ( p ) p n - l ) n - l ~ n , x , - * , n 3 1. (2.1) And, if X # 1 and pqY.cond(x)(here q = 4 ( i f p = 2 ) or p otherwise), for any V m , n € Z , we have L , ( ~ , x ) E L P ( n , x ) ( m o d p ) . (2 .2) For a totally real Abelian number field K of degree n , we have where Rp ( K) is the p-adic regulator of K , D ( K ) is its discriminant, and the sign can be deter- mined canonically: choose orderings in the determinants for Rp ( K ) and so that the Archimedean Rp ( K)/~/D(K) is positve , then use the same orderings in the p-adic case. For a real cyclic sextic field K6 and its quadrtic and cubic subfield K2, K3 , using (2.3) , we have where h - = h ( K 6 ) / h ( K 2 ) h ( K J ) , Rp = R ~ ( K ~ ) / R ~ ( K ~ ) R ~ ( K ~ ) and D - = D ( K 6 ) / ( D ( K 2 ) D(K3) ) . By ref. [ 6 ] , we know that No. 10 CONGRUENCES FOR CLASS NUMBERS 1013 Hence by ( 2 . 1 ) and ( 2 . 2 ) , we have , z l z 5 ( l - ~ n , ( p ) p ~ l - ' ) ( l - ~ ' ~ ~ ( p ) p ~ ~ - ~ ) ( n 1 n 2 ) - ~ ~ n , , ~ ~ , B n ~ , ~ ~ ~ ~ mod^). ( 2 . 6 ) Note that ( 1 - ~ i w - n ( p ) p n - ' ) ~ l ( m o d p ) for n > 1 and V iEZ. ( i ) Suppose that If" . Then p = p i , i = 1 , ... , s . Put n l = e, n 2 = 5 ( p - 1 ) 6 6 ' When p > 7 , by ( 2 . 6 ) , we have 36 4 h R ; / f 6 -- - B e , , , B V , , ~ ~ 5 6 6 ( m o d p ) ; 6 when p = 7 , by ( 2 . 6 ) , we have Let f = cond( X P ) . Then we know that plf; thus p is unramified over the cyclotomic fields 6 Q( C"). f 3 ( i i ) Suppose that p I; and p + 3 . Then p = q j ( j = l;.., t ) . Put n l = e,n2 = f 3 2(p - I ) . By ( 2 . 6 ) , we have 3 Let f = cond ( Xe;l;') . Then we know that p l f ; thus p is unramified over the cyclotomic fields Q(5/.). Note that x'W = ( ~ q ) ~ ) . f 2 ( i i i) Supposethatpl; a n d p + 2 , 3 . T h e n p = r x , k = l , . . . , l . Put n l = n r = y . By f ( 2 . 6 ) , we have 4 h - R ; / f 6 = 4 B p - 1 ,x,! Bp-l , x',! (mod p ) . Let f = cond ( x?) . Then we know that plf. Thus p is unramified over the cyclotomic fields Q( C'). Note that X ' P - ~ = ( X y ) 5 . 2 3 Regulator R, ( K ) By ref. [ 6 ] , we know that, in the relative unit group UR of the real cyclic number field Kg, there exists a relative fundamental unit E such that U, = { + E ~ ( E " ) ' I k , z E z}, and every element in UR can be uniquely represented by * Ek ( E" ) ! In order to calculate R; with the relative fundamental unit E , we will prove the following Lemma 1 first. ~ e m m a 3 . U * = u ~ u ~ u ~ = ( - ~ , E , ~ , ~ ~ , E , E ~ ) i ~ a ~ u b g r o u p o f ~ ~ o f ~ n i t e i n d e x , and 1014 SCIENCE IN CHINA (Series A ) Vol. 42 Q = ( u 6 : u * ) = 12/Q,Q2, ( 3 . 1 ) where Q I = ( U ~ : N ~ / ~ ( U ~ ) ) , Q Z = ( U ~ : ( - ~ ) N ~ / ~ ( U , ) ) ; and, Q l = 1 , 4 ; Q 2 = 1 , 3 ; Q = 1 , 3,4,12. Proof. It is easy to check that Q, = 1 , 4 ; Q2 = 1 , 3 . NOW we only need to prove ( 3 . 1 ) . Con- struct homomorphisms - Here x represents the coset of quotient groups to which x belongs. It is obvious that N3, N2 are well defined and they are homomorphisms. Then ker(N3) = ( x I x E u6, 3 y E U 3 , s . t . N b n ( ~ j ' ) = & 11, ker(N2) = 1; I x E U6, 3 y E Uz, s.t.N6,,(xy) = * 11. Since U * = ( - 1 , E , p , p a , E , E') = U2U3 UR, it is easy to prove that u * / ( - 1 ) = ker (Nz) n ker (K3) . Thus ( u , : u * ) = ( u , / ( - 1 ) : u*/ ( - 1 ) ) = ( u,/(- 1 ) : ker (N3)) (ker (N3) : ker(fi3) n ker(f i2)> = ( ( - l ) N 6 / 3 ( ~ 6 ) : ( - ~ , , U ~ , ( , U ~ ) ~ ) ) ( ~ 6 / 2 ( k e r ( % 3 ) ) : ( - 1 , e 3 ) / ( - 1 ) ) . If we can prove that ~ ~ / ~ ( k e r ( % ~ ) ) = N6/2( ~ 6 ) / ( - I ) , then we have ( u 3 : u * ) = ((- ~ ) N ~ / ~ ( u ~ ) : ( - I , , U ~ , ( , U ~ ) ~ ) ) We end our discussion of Lemma 1 by proving ( 3 . 2 ) . If N6/2(U6) = ( - 1 , c 3 ) , since E € k e r ( N , ) , then ( 3 . 2 ) is obvious; if N ~ / ~ ( u ~ ) # ( - 1 , c 3 ) , then 3 x E U6, N6/2(x) =eB,3Y\w. Note that ( 2 , w ) = 1 , hence Nbl2(U6) = ( - 1 , ~ ) = UZ. Considering that X2€ker (N3) , N ~ / ~ ( x ~ ) = E~~ and ( 2 w , 3 ) = 1, we get N6,(ker(K3)) = u2/( - 1) = N6/2( u 6 ) / ( - 1 ) . This proves ( 3 . 2 ) . 1 By Lemma3, R p ( K 6 ) = g R p ( e , p , P , E , F ) , From ref. [61, we know that V x E UR, N,/, ( x ) = 1 , N6/2 ( x ) = f 1. Then by computation we have Therefore Now we discuss the proof of Lemma 2, namely how to represent the fundmental relative unit. Consider an equation over K , No. 10 CONGRUENCES FOR CLASS NUMBERS 1015 2 whose roots are 6 , 8" , Bb , here f3 = ( a2 + 3 b 2 ) / 4 [ a G 2 , b 0 ( m o d 3 ) , b > 0 , 3 ) l f 3 ; and f3 is uniquely represented by the above formulae (see sec. 2 of ref . [ 6 ] ) . Note that V p I f 3 , p ' 1 2 3 . Then ( 3 . 4 ) is Eisenstein; thus v p ( B ) = up(@') = up(&' ) = -. Consider 3 where xi , yi , i = 0 , 1 , 2 , are integers except 2 ,3 (see Theorem 1 of reference [ 6 ] ) . 2 L e t r = r ( ~ ) = ~ ~ ( n ) q b e ~ a u s s s u m . B y r e f . [ 6 ] , w e h a v e & = f ~ , r = f ~ # , t t - I . ( n . f , ) = l 1 6 0 " I ea 8"' 1 p2 e and = - ( e + P + e a 2 ) 2 + 3 ( e ~ + B"B"'+ ~ " ' 6 ) =-f3. where p=e2?= - ,$= ( - 1) k a + 6 6 3 2 2 ; k , m are the integers uniquely determined Since 19 is an integer except 3 , { 1 ,6,8" } is an p-integer basis for p 3 . It is not difficult to prove that Q Y E O6 , ( integer domain of K , ) , - ( - l ) m ( r + ~ ) , f= - ( - 1 ) " by f3(ref. [ 6 ] ) . Therefore 0 = - 3 ( - l ) " ( p 2 T + p 2 r ) , @ =- 3 3 ( P + - p r ) . Then we can write x18 + x2B" and y l0 + y20a in ( 3 . 5 ) as - - ~ , e + ~ ~ 8 " = y r + y r , y l e + ~ ~ 8 6 = ylr + Y I T , ( 3 . 6 ) where y , y' E & ( m) are integers except 2 , 3 . Now note that N6/3 ( E) = 1. We can write where x = xo , X I = yo E 5, y E Q( G) are integers except 2 , 3 , and Thus we arrives at the conclusion of Lemma 2. 1016 SCIENCE IN CHINA (Series A ) Vol. 42 Now let us compute the congruences of R , /f6 in three cases. ( i ) Suppose that p = p i , i = 1 , ..., s . Since up ( E ) = 0, we have up 1 1 ($$(yo+ y , 8 + y 2 B ) ) ~ o . Note that up = - 2. Then vp(yo + y,B + y 2 B ) 3 y. Futhermore , since yo + y ,8 + y,B" € K3 and p is purely ramified over K3, we have up ( yo + y1 8 + 2 1 y20") 3 7; thus p I yo and vp 8 6; that is, p I ( x 2 - I ) , - 1 1 - ~ , ( 2 / ( x + y r + y r ) ~ - l ) ~ ~ . Put 1 - i = cp, c E Q , U = ~ E Q ( ~ ~ ) , A = u r + u r , T X X = A + 2/ cp + 2A + A'. Noting that up( T) 3 1/6, we have - = (log,x + log,(l + ur + ur + Jcp + 2 ( u r + G) + ( u r + G ) 2 ) ) 2 If p > 7 , we have T2 r3 T~ T5 ( l o g , ~ ) 2 = T - - + - - - ( 2 3 4 ' 7 - 6 3 Because of symmetry of R; = - log,^"')^, we can omit the terms in the expansion of T whose Q i = o A i ( d c p + 2 A +A2) j and jareodds. Since v p ( A ) a 1 / 3 , wecanalsoomit A', i a4 in theexpan- sion of T k. Through complicated computation, we have From = f3 , r3 = f 3 # we get 7 If p = 7 , (logpE) = (mod?%). Using similar tricks, we have 3 2 - 3 - R ; - - ( - ~7 + - u u ~ , - ( - 1 ) ( ~ 3 + + i 3 a ) f 3 - ,(m)2fi Q 3 - 0 2 ( i i) Suppose t h a t p = q j , j = l , . . . , t . By ( 3 . 7 ) , wecanwrite E = x + y r + y r + , - ( x l + f = qj. j = l , . . . , t , weknowthat Xi, i = 0 , 1 , areallp-integers; thatis, vp (Xi )aO, v i = 0 , 1 . By ( 3 . 8 ) , we have No. 10 CONGRUENCES FOR CLASS NUMBERS 1017 Squaring both sides of ( 3 . 9 ) , we get f 2 f 2 x2 - 1 + 2& = ( x t 2 + 2y1 7 f 3 ) - 2xy + y2# = (2x t yT + yI2#) -. (3 .10) f K 2 ' f C 2 1 4 Note that N6/2 ( E ) = EE" Eo = + 1. Then xi - 3 f 3 x o x , x , + f3(+x: + 3%:) = * 1. 1 ~ h u s x:= rt 1 ( m o d p ) . Put T = ( X l r +=)/X,. Since v,(T)>-, we get 3 Using symmetry of R p and tricks similar to those in ( i ) , we then obtain Finally, by (3.10) and X;= * l ( m o d p ) , we have - f 3 f 2 4 + 8 ( y y ' + yyt)-) ( m o d p i ) . f X 2 (3.11) - If Xo= + l ( m o d p ) , then E = X o + X l s + X , r x 0 r * l ( m o d p ) ; thus x + y r + y r + 1 1 2 / ( x + y r + F ) 2 - 1 - * l ( m o d p ? . ) , x + I + 2 / ~ 1 r 0 ( m o d p ? ) , ( x * 1 ) 2 = ~ 2 - 1 1 0 2 ( m o d p ? ) , x z + l ( m o d p ) , Furthermore, since X o = x +,-x'r * l ( m o d p ) , we have ,-x' f f r O ( m o d p ) ; that is, p I x l . Combining this with (3.11) and x3= * l ( m o d p ) , we get ( i i i ) Suppose that p = r k , k = 1 , I . Then v, ( J ( x + F + ~ ) 2 - l ) = vp have 3 log,^ sz ( log ( 1 + F1))2(modpi) L L~ - 1 3 - - - L~ (mod p ? ) 3 L' - 1 (mod pf), 3 3 3 2 3 R - (L' - 1 ) (mod p i ) -- -6(x - 1 + 2@f3) (mod pY). Q ,=, Q 4 Proof of the theorems and examples Based on the results on the regulators in sec. 3 and the formulae in sec. 2, through computa- 1018 SCIENCE IN CHINA (Series A ) Vol. 42 tion, we can obtain the theorems at the beginning of this paper. The proof of those theorems are obvi- ous except one point : on the right side of each congruence there are Be, Bre, x; E Q( Ci) , where j is the conductor of X P = Xct) - " . Hy sec . 2 , we know that p is unramified over Q( Ci). So we can sub- 1 1 1 stitute (mod , 6 = , - , - , with (mod p ) on the right side of each congruence. 6 3 2 I f p = q j , j = l ; - - , t , we knowthat X ; = + l ( m o d p ) . Soweconjecturethat X o = *l(rnod p ) at least in most cases, where the formula is much simpler. Example 1. Let ,f6 = 13 be the conductor of real cyclic sextic number field K6. By the ap- 1 n 3 pendixofref. [ 6 ] , we know that E = - ( - 1 + 2 1 9 + 2 8 " ) + - ( 3 + 2 8 - 6 8 " ) = x + yr+F+ 2 13 - 7 ~ ' ( x + ~ i + ~ ) ~ - l , where x = - - , 6 Y = ? , p = . Note that x = l ( m o d 1 3 ) . 2 With the appendix of ref. [ 6 ] , we can calculate out Q = 12, # = - ( 5 + 3 C 3 ) 2 , B d = B 2 = 6 1 5 - B S ( P - 1 ) = B I 0 = M) . By Theorem 1 , we get h - E 1 (mod 13) . This result is identified with that 6 ' 6 ofref. [6 ] whichshows that h - = I . Hence h ( K g ) = h - ( K g ) = l . Acknowledgment The author is grateful to his tutor Prof. Zhang Xianke for his guide and help in the seminar of algebraic n u d e r theory and arithmetic algebraic geometry. References Ankeny, N . , Atrin E . , Chowla, S . , The class number of real quadratic number fields, Ann. of Math. , 1952, 52(2) : 479. Lu , H . W . , Congruences for the class number of qudratic fields, Abb . Math. Sen. Univ . Hamburg, 1982, 52 : 254. Feng, K. Q . , Ankeny- Artin-Chowla formula on cubic cyclic n u d e r fields, J. 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